How do you solve $log_2c>8$ ?

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Answer 1 · 2016-12-29T04:19:21

$c>256$

The domain of $c$ in $log_2c$ is $c>0$

and then the derivative of $log_2c$ , which is $1/(cln2)$ , too is positive.

Hence $log_2c$ is a monotonically increasing function of $c$ .

Hence, as $log_2c>8$ , $c>2^8$

or $c>256$
graph{(y-lnx/ln2)(y-8)=0 [-50, 300, -2, 10]}

How do you solve log_2c>8log_2c>8?